Is there a relationship between Rotors and the Rodrigues' rotation formula

Question

I am trying to understand quaternion in general, and it seems like the path to making sense of how they actually work is to first understand rotors and other techniques related to rotations. By looking at the equations and just reading briefly about these topics, it seems like there are relationships between all these topics: rotors, the Rodrigues' rotation formula which involves a scalar and a vector product as well as quaternions. Before getting on to the topic of quaternions all I am trying to do now, is finding out if there is any sort of connection between rotors (or more precisely using rotors to rotate a vector) as the equation to rotate a vector using a rotor also involves scalar and vector product, and the Rodrigues' formula:

$\mathbf{v}_\mathrm{rot} = \mathbf{v} \cos\theta + (\mathbf{k} \times \mathbf{v})\sin\theta + \mathbf{k} (\mathbf{k} \cdot \mathbf{v}) (1 - \cos\theta)$

If so, could someone explain it please?

Answer 1

IMHO the set of quaternions is more appropriately viewed as a scalar bivector pair. The rotation equation is a similarity transform. It's certainly reasonable to take a GA rotor view. The best way to think about it is going to depend on what's more familiar to you.

Answer 2

We could "derive" the Rodrigues formula by "vectorizing" the quaternion sandwitch product. First, let us recall the sadwitch product:

$v' = Q \ v \ Q^*$

Where $v$ is a pure quaternion (its real part equal to zero) and $Q$ is a unit quaternion and $Q^*$ is its conjugate:

$Q^{*} = q_0 - i q_1 - j q_2 - k q_3$

Aplying the Euler relation:

$v' = Q \ v \ Q^{-1} = (\cos \frac{\theta}{2} + \sin \frac{\theta}{2} \ b) \ v \ (\cos \frac{\theta}{2} - \sin \frac{\theta}{2} \ b)$

$v' = (Q \ v) \ Q^{-1} = (\cos \frac{\theta}{2} \ v + \sin \frac{\theta}{2} \ b \ v) \ (\cos \frac{\theta}{2} - \sin \frac{\theta}{2} \ b)$

$v' = \cos^2 \frac{\theta}{2} \ v - (\cos \frac{\theta}{2} \ \sin \frac{\theta}{2}) \ v \ b + (\sin \frac{\theta}{2} \ \cos \frac{\theta}{2}) \ b \ v - \sin^2 \frac{\theta}{2} \ b \ v \ b$

The product $v \ b$ that apperars in the above expressions is the product of two pure quaternions and it is defined in vector form as:

$v \ b = (i v_0 + j v_1 + k v_2) \ (i b_0 + j b_1 + k b_2) = -v \cdot b + v \times b$,

where $v \cdot b$ is the dot product and $v \times b$ is the cross product.

Notice that the algebraic sum $-v \ b + b \ v$ is not zero, since the cross product is non-commutative. As a consequence $-v \ b + b \ v = (v \cdot b - v \times b) + (-b \cdot v + b \times v) = 0 + 2 \ b \times v = 0 - 2 \ v \times b$.

Replacing that into our last equation we get:

$v' = \cos^2 \frac{\theta}{2} \ v + 2 \ (\cos \frac{\theta}{2} \ \sin \frac{\theta}{2}) \ b \times v - \sin^2 \frac{\theta}{2} \ b \ v \ b$

Interestingly the expression $b \ v \ b$ is nothing more than the reflection of $v$ in the plane with normal $b$. Where $b$ and $v$ are pure quaternions. Reflection can be expressed in vector form as:

$b \ v \ b = 2 \ (-v \cdot b) \ b + v$

where the vector $v$ is translated in direction of the negative normal $b$ twice the distance of $v$ projected in $b$. Finally the equation is:

$v' = \cos^2 \frac{\theta}{2} \ v + 2 \ (\cos \frac{\theta}{2} \ \sin \frac{\theta}{2}) \ b \times v - \sin^2 \frac{\theta}{2} \ (2 \ (-v \cdot b) \ b + v)$

$v' = \cos^2 \frac{\theta}{2} \ v - \sin^2 \frac{\theta}{2} \ v + 2 \ (\cos \frac{\theta}{2} \ \sin \frac{\theta}{2}) \ b \times v + 2 \ \sin^2 \frac{\theta}{2} \ (v \cdot b) \ b$

Applying the following trigonometric identities:

$\cos \theta = \cos^2 \frac{\theta}{2} - \sin^2 \frac{\theta}{2}$

$\sin \theta = 2 \cos \frac{\theta}{2} \ \sin \frac{\theta}{2}$

$1 - \cos \theta = 2 \ \sin^2 \frac{\theta}{2}$

We get the Rodrigues formula:

$v' = \cos \theta \ v + \sin \theta \ b \times v + (1 - \cos \theta) \ ( v \cdot b ) \ b $

It was discovered by Olinde Rodriguez three years before Hamilton discover quaternions.